Question

[DD]

Rút gọn biểu thức:

P = \(\dfrac{2a^2+4}{1-a^3}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}\)

A = \(\dfrac{x}{x-1}+\dfrac{3}{x+1}-\dfrac{6x-4}{x^2-1}\)

Answer 1

[Đoàn Trần Quỳnh Hương]

ĐKXĐ: a > 0 và a khác 1

\(P=\dfrac{2\left(a^2+2\right)}{\left(1-a\right)\left(1+a+a^2\right)}-\dfrac{\left(1-\sqrt{a}\right)\left(1+a+a^2\right)}{\left(1-a\right)\left(1+a+a^2\right)}-\dfrac{\left(1+\sqrt{a}\right)\left(1+a+a^2\right)}{\left(1-a\right)\left(1+a+a^2\right)}\)\(=\dfrac{2a^2+4-\left(1+a+a^2\right).\left(1-\sqrt{a}+1+\sqrt{a}\right)}{1-a^3}\)

\(=\dfrac{2a^2+4-\left(1+a+a^2\right)}{1-a^3}\)

\(=\dfrac{a^2+a+3}{\left(1-a^3\right)}\)